Swift 3.0 & Swift 4.0 & Swift 5.0中最简单的解决方案

func delayWithSeconds(_ seconds: Double, completion: @escaping () -> ()) {
    DispatchQueue.main.asyncAfter(deadline: .now() + seconds) { 
        completion()
    }
}

使用

delayWithSeconds(1) {
   //Do something
}

1)添加这个方法作为UIViewController Extension的一部分。

extension UIViewController{
func runAfterDelay(delay: NSTimeInterval, block: dispatch_block_t) {
        let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
        dispatch_after(time, dispatch_get_main_queue(), block)
    }
}

在VC上调用这个方法:

    self.runAfterDelay(5.0, block: {
     //Add code to this block
        print("run After Delay Success")
    })

2)

performSelector("yourMethod Name", withObject: nil, afterDelay: 1)

3)

override func viewWillAppear(animated: Bool) {

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, 2), dispatch_get_main_queue(), { () -> () in
    //Code Here
})

/ /紧凑的形式

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, 2), dispatch_get_main_queue()) {
    //Code here
 }
}

这对我很管用。

斯威夫特3:

let time1 = 8.23
let time2 = 3.42

// Delay 2 seconds

DispatchQueue.main.asyncAfter(deadline: .now() + 2.0) {
    print("Sum of times: \(time1 + time2)")
}

objective - c:

CGFloat time1 = 3.49;
CGFloat time2 = 8.13;

// Delay 2 seconds

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(2.0 * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
    CGFloat newTime = time1 + time2;
    NSLog(@"New time: %f", newTime);
});

保留当前队列!

除了很好地回答这个问题之外，您还可以考虑保留当前队列以防止不必要的主队列操作(例如，当您试图延迟一些异步操作时)。

func after(_ delay: TimeInterval,
           perform block: @escaping ()->(),
           on queue: DispatchQueue = OperationQueue.current?.underlyingQueue ?? .main) { // So this `queue` preserves the current queue and defaulted to the `main`. Also the caller can pass in the desired queue explicitly
    queue.asyncAfter(deadline: .now() + delay, execute: block)
}

用法:

after(3) {
    // will be executed on the caller's queue
    print(Date())
}

使用这段代码在2.0秒后执行一些UI相关的任务。

            let delay = 2.0
            let delayInNanoSeconds = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
            let mainQueue = dispatch_get_main_queue()

            dispatch_after(delayInNanoSeconds, mainQueue, {

                print("Some UI related task after delay")
            })

Swift 3.0版本

以下闭包函数在主线程上执行一些延迟后的任务。

func performAfterDelay(delay : Double, onCompletion: @escaping() -> Void){

    DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + delay, execute: {
       onCompletion()
    })
}

像这样调用这个函数:

performAfterDelay(delay: 4.0) {
  print("test")
}

斯威夫特 3+

这在Swift 3+中是超级简单和优雅的:

DispatchQueue.main.asyncAfter(deadline: .now() + 4.5) {
    // ...
}

年长的回答:

为了扩展Cezary的答案，它将在1纳秒后执行，我必须执行以下操作以在4秒半后执行。

let delay = 4.5 * Double(NSEC_PER_SEC)
let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay))
dispatch_after(time, dispatch_get_main_queue(), block)

编辑:我发现我原来的代码有一点错误。如果不将NSEC_PER_SEC转换为Double类型，隐式类型将导致编译错误。

如果有人能提出一个更优的解决方案，我很乐意听听。